`vecb and vecc` are non- collinear if `veca xx (vecb xx vecc) + (veca .vecb) vecb = ( 4-2x- sin y) vecb + ( x^(2) -1) vecc nad (veca. vecc) veca =veca ` then a. x =1 b. x = -1 c. `y = (4 n+1) pi/2, n in I ` d. `y ( 2n + 1) pi/2, n in I`

Options

A. x =1
B. x = -1
C. `y = (4 n+1) pi/2, n in I `
D. `y ( 2n + 1) pi/2, n in I`

Explanation

To solve the given problem step by step, we will break down the equations involving vectors and analyze them systematically.

### Step 1: Write down the given equation
We start with the equation:
\[ \vec{a} \times (\vec{b} \times \vec{c}) + (\vec{a} \cdot \vec{b}) \vec{b} = (4 - 2x - \sin y) \vec{b} + (x^2 - 1) \vec{c} \]

### Step 2: Use the vector triple product identity
Using the vector triple product identity, we can rewrite \(\vec{a} \times (\vec{b} \times \vec{c})\) as:
\[ \vec{a} \cdot \vec{c} \vec{b} - \vec{a} \cdot \vec{b} \vec{c} \]
Thus, substituting this into our equation gives:
\[ (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} + (\vec{a} \cdot \vec{b}) \vec{b} = (4 - 2x - \sin y) \vec{b} + (x^2 - 1) \vec{c} \]

### Step 3: Rearranging the equation
We can rearrange the equation:
\[ (\vec{a} \cdot \vec{c} + \vec{a} \cdot \vec{b}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} = (4 - 2x - \sin y) \vec{b} + (x^2 - 1) \vec{c} \]

### Step 4: Equate coefficients of \(\vec{b}\) and \(\vec{c}\)
From the equation, we can equate the coefficients of \(\vec{b}\) and \(\vec{c}\):
1. For \(\vec{b}\):
\[ \vec{a} \cdot \vec{c} + \vec{a} \cdot \vec{b} = 4 - 2x - \sin y \quad \text{(Equation 1)} \]
2. For \(\vec{c}\):
\[ -\vec{a} \cdot \vec{b} = x^2 - 1 \quad \text{(Equation 2)} \]

### Step 5: Solve Equation 2 for \(\vec{a} \cdot \vec{b}\)
From Equation 2:
\[ \vec{a} \cdot \vec{b} = 1 - x^2 \]

### Step 6: Substitute into Equation 1
Substituting \(\vec{a} \cdot \vec{b}\) into Equation 1:
\[ \vec{a} \cdot \vec{c} + (1 - x^2) = 4 - 2x - \sin y \]
This simplifies to:
\[ \vec{a} \cdot \vec{c} = 4 - 2x - \sin y - 1 + x^2 \]
\[ \vec{a} \cdot \vec{c} = 3 - 2x - \sin y + x^2 \quad \text{(Equation 3)} \]

### Step 7: Use the condition \(\vec{a} \cdot \vec{c} \vec{a} = \vec{a}\)
We are given that:
\[ \vec{a} \cdot \vec{c} \vec{a} = \vec{a} \]
This implies:
\[ \vec{a} \cdot \vec{c} = 1 \]
Substituting this into Equation 3 gives:
\[ 1 = 3 - 2x - \sin y + x^2 \]
Rearranging this results in:
\[ 0 = x^2 - 2x + 2 - \sin y \]
\[ \sin y = x^2 - 2x + 2 \quad \text{(Equation 4)} \]

### Step 8: Analyze the equation for \(\sin y\)
The expression \(x^2 - 2x + 2\) can be rewritten as:
\[ \sin y = (x - 1)^2 + 1 \]
Since \(\sin y\) must be in the range \([-1, 1]\), we see that \((x - 1)^2 + 1\) is always greater than or equal to 1. Thus, for \(\sin y\) to be valid, we must have:
\[ (x - 1)^2 + 1 \leq 1 \]
This implies:
\[ (x - 1)^2 = 0 \Rightarrow x = 1 \]

### Step 9: Substitute \(x = 1\) into Equation 4
Substituting \(x = 1\) into Equation 4 gives:
\[ \sin y = 1 \]
Thus, \(y\) can take the values:
\[ y = \frac{\pi}{2} + 2n\pi \quad \text{for } n \in \mathbb{Z} \]
This can also be expressed as:
\[ y = (4n + 1)\frac{\pi}{2} \quad \text{for } n \in \mathbb{Z} \]

### Conclusion
The correct answers are:
- \(x = 1\) (Option A)
- \(y = (4n + 1)\frac{\pi}{2}, n \in \mathbb{Z}\) (Option C)